# 9.3 Multiplication of square root expressions

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The distinction between the principal square root of the number x and the secondary square root of the number x is made by explanation and by example. The simplification of the radical expressions that both involve and do not involve fractions is shown in many detailed examples; this is followed by an explanation of how and why radicals are eliminated from the denominator of a radical expression. Real-life applications of radical equations have been included, such as problems involving daily output, daily sales, electronic resonance frequency, and kinetic energy.Objectives of this module: be able to use the product property of square roots to multiply square roots.

## Overview

• The Product Property of Square Roots
• Multiplication Rule for Square Root Expressions

## The product property of square roots

In our work with simplifying square root expressions, we noted that

$\sqrt{xy}=\sqrt{x}\sqrt{y}$

Since this is an equation, we may write it as

$\sqrt{x}\sqrt{y}=\sqrt{xy}$

To multiply two square root expressions, we use the product property of square roots.

## The product property $\sqrt{x}\sqrt{y}=\sqrt{xy}$

$\sqrt{x}\sqrt{y}=\sqrt{xy}$

The product of the square roots is the square root of the product.

In practice, it is usually easier to simplify the square root expressions before actually performing the multiplication. To see this, consider the following product:

$\sqrt{8}\sqrt{48}$
We can multiply these square roots in either of two ways:

Simplify then multiply.

$\sqrt{4·2}\sqrt{16·3}=\left(2\sqrt{2}\right)\left(4\sqrt{3}\right)=2·4\sqrt{2·3}=8\sqrt{6}$

Multiply then simplify.

$\sqrt{8}\sqrt{48}=\sqrt{8·48}=\sqrt{384}=\sqrt{64·6}=8\sqrt{6}$

Notice that in the second method, the expanded term (the third expression, $\sqrt{384}$ ) may be difficult to factor into a perfect square and some other number.

## Multiplication rule for square root expressions

The preceding example suggests that the following rule for multiplying two square root expressions.

## Rule for multiplying square root expressions

1. Simplify each square root expression, if necessary.
2. Perform the multiplecation.
3. Simplify, if necessary.

## Sample set a

Find each of the following products.

$\sqrt{3}\sqrt{6}=\sqrt{3·6}=\sqrt{18}=\sqrt{9·2}=3\sqrt{2}$

$\sqrt{8}\sqrt{2}=2\sqrt{2}\sqrt{2}=2\sqrt{2·2}=2\sqrt{4}=2·2=4$

This product might be easier if we were to multiply first and then simplify.

$\sqrt{8}\sqrt{2}=\sqrt{8·2}=\sqrt{16}=4$

$\sqrt{20}\sqrt{7}=\sqrt{4}\sqrt{5}\sqrt{7}=2\sqrt{5·7}=2\sqrt{35}$

$\begin{array}{lll}\sqrt{5{a}^{3}}\sqrt{27{a}^{5}}=\left(a\sqrt{5a}\right)\left(3{a}^{2}\sqrt{3a}\right)\hfill & =\hfill & 3{a}^{3}\sqrt{15{a}^{2}}\hfill \\ \hfill & =\hfill & 3{a}^{3}·a\sqrt{15}\hfill \\ \hfill & =\hfill & 3{a}^{4}\sqrt{15}\hfill \end{array}$

$\begin{array}{lll}\sqrt{{\left(x+2\right)}^{7}}\sqrt{x-1}=\sqrt{{\left(x+2\right)}^{6}\left(x+2\right)}\sqrt{x-1}\hfill & =\hfill & {\left(x+2\right)}^{3}\sqrt{\left(x+2\right)}\sqrt{x-1}\hfill \\ \hfill & =\hfill & {\left(x+2\right)}^{3}\sqrt{\left(x+2\right)\left(x-1\right)}\hfill \\ \begin{array}{cccc}& & & \begin{array}{cccc}& & & \begin{array}{cccc}& & & \text{or}\end{array}\end{array}\end{array}\hfill & =\hfill & {\left(x+2\right)}^{3}\sqrt{{x}^{2}+x-2}\hfill \end{array}$

## Practice set a

Find each of the following products.

$\sqrt{5}\sqrt{6}$

$\sqrt{30}$

$\sqrt{32}\sqrt{2}$

8

$\sqrt{x+4}\sqrt{x+3}$

$\sqrt{\left(x+4\right)\left(x+3\right)}$

$\sqrt{8{m}^{5}n}\sqrt{20{m}^{2}n}$

$4{m}^{3}n\sqrt{10m}$

$\sqrt{9{\left(k-6\right)}^{3}}\sqrt{{k}^{2}-12k+36}$

$3{\left(k-6\right)}^{2}\sqrt{k-6}$

$\sqrt{3}\left(\sqrt{2}+\sqrt{5}\right)$

$\sqrt{6}+\sqrt{15}$

$\sqrt{2a}\left(\sqrt{5a}-\sqrt{8{a}^{3}}\right)$

$a\sqrt{10}-4{a}^{2}$

$\sqrt{32{m}^{5}{n}^{8}}\left(\sqrt{2m{n}^{2}}-\sqrt{10{n}^{7}}\right)$

$8{m}^{3}{n}^{2}\sqrt{n}-8{m}^{2}{n}^{5}\sqrt{5m}$

## Exercises

$\sqrt{2}\sqrt{10}$

$2\sqrt{5}$

$\sqrt{3}\sqrt{15}$

$\sqrt{7}\sqrt{8}$

$2\sqrt{14}$

$\sqrt{20}\sqrt{3}$

$\sqrt{32}\sqrt{27}$

$12\sqrt{6}$

$\sqrt{45}\sqrt{50}$

$\sqrt{5}\sqrt{5}$

5

$\sqrt{7}\sqrt{7}$

$\sqrt{8}\sqrt{8}$

8

$\sqrt{15}\sqrt{15}$

$\sqrt{48}\sqrt{27}$

36

$\sqrt{80}\sqrt{20}$

$\sqrt{5}\sqrt{m}$

$\sqrt{5m}$

$\sqrt{7}\sqrt{a}$

$\sqrt{6}\sqrt{m}$

$\sqrt{6m}$

$\sqrt{10}\sqrt{h}$

$\sqrt{20}\sqrt{a}$

$2\sqrt{5a}$

$\sqrt{48}\sqrt{x}$

$\sqrt{75}\sqrt{y}$

$5\sqrt{3y}$

$\sqrt{200}\sqrt{m}$

$\sqrt{a}\sqrt{a}$

$a$

$\sqrt{x}\sqrt{x}$

$\sqrt{y}\sqrt{y}$

$y$

$\sqrt{h}\sqrt{h}$

$\sqrt{3}\sqrt{3}$

3

$\sqrt{6}\sqrt{6}$

$\sqrt{k}\sqrt{k}$

$k$

$\sqrt{m}\sqrt{m}$

$\sqrt{{m}^{2}}\sqrt{m}$

$m\sqrt{m}$

$\sqrt{{a}^{2}}\sqrt{a}$

$\sqrt{{x}^{3}}\sqrt{x}$

${x}^{2}$

$\sqrt{{y}^{3}}\sqrt{y}$

$\sqrt{y}\sqrt{{y}^{4}}$

${y}^{2}\sqrt{y}$

$\sqrt{k}\sqrt{{k}^{6}}$

$\sqrt{{a}^{3}}\sqrt{{a}^{5}}$

${a}^{4}$

$\sqrt{{x}^{3}}\sqrt{{x}^{7}}$

$\sqrt{{x}^{9}}\sqrt{{x}^{3}}$

${x}^{6}$

$\sqrt{{y}^{7}}\sqrt{{y}^{9}}$

$\sqrt{{y}^{3}}\sqrt{{y}^{4}}$

${y}^{3}\sqrt{y}$

$\sqrt{{x}^{8}}\sqrt{{x}^{5}}$

$\sqrt{x+2}\sqrt{x-3}$

$\sqrt{\left(x+2\right)\left(x-3\right)}$

$\sqrt{a-6}\sqrt{a+1}$

$\sqrt{y+3}\sqrt{y-2}$

$\sqrt{\left(y+3\right)\left(y-2\right)}$

$\sqrt{h+1}\sqrt{h-1}$

$\sqrt{x+9}\sqrt{{\left(x+9\right)}^{2}}$

$\left(x+9\right)\sqrt{x+9}$

$\sqrt{y-3}\sqrt{{\left(y-3\right)}^{5}}$

$\sqrt{3{a}^{2}}\sqrt{15{a}^{3}}$

$3{a}^{2\text{\hspace{0.17em}}}\sqrt{5a}$

$\sqrt{2{m}^{4}{n}^{3}}\sqrt{14{m}^{5}n}$

$\sqrt{12{\left(p-q\right)}^{3}}\sqrt{3{\left(p-q\right)}^{5}}$

$6{\left(p-q\right)}^{4}$

$\sqrt{15{a}^{2}{\left(b+4\right)}^{4}}\sqrt{21{a}^{3}{\left(b+4\right)}^{5}}$

$\sqrt{125{m}^{5}{n}^{4}{r}^{8}}\sqrt{8{m}^{6}r}$

$10{m}^{5}{n}^{2}{r}^{4}\sqrt{10mr}$

$\sqrt{7{\left(2k-1\right)}^{11}{\left(k+1\right)}^{3}}\sqrt{14{\left(2k-1\right)}^{10}}$

$\sqrt{{y}^{3}}\sqrt{{y}^{5}}\sqrt{{y}^{2}}$

${y}^{5}$

$\sqrt{{x}^{6}}\sqrt{{x}^{2}}\sqrt{{x}^{9}}$

$\sqrt{2{a}^{4}}\sqrt{5{a}^{3}}\sqrt{2{a}^{7}}$

$2{a}^{7}\sqrt{5}$

$\sqrt{{x}^{n}}\sqrt{{x}^{n}}$

$\sqrt{{y}^{2}n}\sqrt{{y}^{4}n}$

${y}^{3n}$

$\sqrt{{a}^{2n+5}}\sqrt{{a}^{3}}$

$\sqrt{2{m}^{3n+1}}\sqrt{10{m}^{n+3}}$

$2{m}^{2n+2}\sqrt{5}$

$\sqrt{75{\left(a-2\right)}^{7}}\sqrt{48a-96}$

$\sqrt{2}\left(\sqrt{8}+\sqrt{6}\right)$

$2\left(2+\sqrt{3}\right)$

$\sqrt{5}\left(\sqrt{3}+\sqrt{7}\right)$

$\sqrt{3}\left(\sqrt{x}+\sqrt{2}\right)$

$\sqrt{3x}+\sqrt{6}$

$\sqrt{11}\left(\sqrt{y}+\sqrt{3}\right)$

$\sqrt{8}\left(\sqrt{a}-\sqrt{3a}\right)$

$2\sqrt{2a}-2\sqrt{6a}$

$\sqrt{x}\left(\sqrt{{x}^{3}}-\sqrt{2{x}^{4}}\right)$

$\sqrt{y}\left(\sqrt{{y}^{5}}+\sqrt{3{y}^{3}}\right)$

${y}^{2}\left(y+\sqrt{3}\right)$

$\sqrt{8{a}^{5}}\left(\sqrt{2a}-\sqrt{6{a}^{11}}\right)$

$\sqrt{12{m}^{3}}\left(\sqrt{6{m}^{7}}-\sqrt{3m}\right)$

$6{m}^{2}\left({m}^{3}\sqrt{2}-1\right)$

$\sqrt{5{x}^{4}{y}^{3}}\left(\sqrt{8xy}-5\sqrt{7x}\right)$

## Exercises for review

( [link] ) Factor ${a}^{4}{y}^{4}-25{w}^{2}.$

$\left({a}^{2}{y}^{2}+5w\right)\left({a}^{2}{y}^{2}-5w\right)$

( [link] ) Find the slope of the line that passes through the points $\left(-5,4\right)$ and $\left(-3,4\right).$

( [link] ) Perform the indicated operations:

$\frac{15{x}^{2}-20x}{6{x}^{2}+x-12}·\frac{8x+12}{{x}^{2}-2x-15}÷\frac{5{x}^{2}+15x}{{x}^{2}-25}$

$\frac{4\left(x+5\right)}{{\left(x+3\right)}^{2}}$

( [link] ) Simplify $\sqrt{{x}^{4}{y}^{2}{z}^{6}}$ by removing the radical sign.

( [link] ) Simplify $\sqrt{12{x}^{3}{y}^{5}{z}^{8}}.$

$2x{y}^{2}{z}^{4}\sqrt{3xy}$

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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